SYSTEMATIC MATHEMATICS

Galois Theory & Finite Fields

Why can polynomial roots be permuted only in certain ways, when does a subgroup determine an intermediate field, and why do some equations admit radicals while the general quintic does not? This course answers those questions in forty-six visible bilingual chapters across nine content-sized units of lengths 5, 5, 6, 5, 5, 5, 6, 5, and 4. It builds field extensions, splitting fields, separability, normality, finite fields, Galois groups and fixed fields, the fundamental correspondence, norms and traces, solvability by radicals, cyclotomic fields, constructibility, Frobenius evidence, exact computational audits, and one complete x^4−2 dossier. Every base field, irreducibility claim, root image, degree, subgroup, fixed element, and numerical-versus-exact boundary is named. The course does not infer a Galois group from one factorization, treat a computer algebra output as proof, or claim to replace commutative algebra, algebraic number theory, infinite Galois theory, or arithmetic geometry.

Before this course: Completed Abstract Algebra, Number Theory, Proof, Logic & Set Theory, and Linear Algebra. Students should already know fields, ideals, quotient rings, polynomial division, irreducibility criteria, groups, normal subgroups, quotient groups, group actions, vector-space bases, determinants, and modular arithmetic. No computer algebra system, commutative algebra course, algebraic closure construction, category theory, or algebraic number theory is assumed.

COURSE FACTSLevel, chapters, units, prerequisite, and outcome
Chapter 1

A field extension is also a vector space whose degree measures algebraic size

Objective: Why is {1,√2,2} not a basis even though it spans the same field?

Writing F⊂E means the operations and identities agree, so E can be viewed as a vector space over F. Its dimension records how many F-coordinates are needed, not how many set elements the fields contain. Begin by naming the base field, extension field, polynomial ring, chosen root, degree, and whether the claim concerns an embedding, automorphism, or abstract isomorphism. The exact result is: If E/F has finite degree n, multiplication by any fixed x∈E is an F-linear map on an n-dimensional vector space, and every element of E is algebraic over F.

The object being calculated is: A field extension E/F is an inclusion of fields with F a subfield of E. Its degree is [E:F]=dim_F E, finite or infinite. A basis {e_i} gives every x∈E one unique finite expression x=Σa_ie_i with a_i∈F. Galois theory turns polynomial symmetry into group symmetry only after existence, separability, normality, and fixed-field hypotheses are checked.

Rebuild the exact problem “Show that Q(√2) has Q-basis {1,√2} and compute (3+2√2)^{-1} in that basis.” without copying a field diagram. Verify every irreducibility, degree, root image, and fixed element, then test this limitation: Finite degree implies algebraic, but an algebraic extension can have infinite degree when it requires infinitely many independent algebraic generators.

If E/F has finite degree n, multiplication by any fixed x∈E is an F-linear map on an n-dimensional vector space, and every element of E is algebraic over F.

The n+1 vectors 1,x,x²,…,x^n lie in the n-dimensional F-space E and are linearly dependent.

Thus coefficients a_0,…,a_n∈F, not all zero, satisfy a_0+a_1x+…+a_nx^n=0.

The nonzero polynomial Σa_it^i has x as a root, which is exactly algebraicity over F.

Show that Q(√2) has Q-basis {1,√2} and compute (3+2√2)^{-1} in that basis.

  1. Every rational expression in √2 reduces using (√2)²=2 to a+b√2, so {1,√2} spans.
  2. If a+b√2=0 with rational a,b and b≠0, then √2=−a/b would be rational; hence a=b=0 and the pair is independent.
  3. Multiply by the conjugate: 1/(3+2√2)=(3−2√2)/(9−8)=3−2√2.

Result: [Q(√2):Q]=2 with basis {1,√2}, and (3+2√2)^{-1}=3−2√2.